A car moving \( 18.2 \mathrm{~m} / \mathrm{s} \) starts to coast UP a frictionless \( 10.0^{\circ} \) hill. How far does it roll before coming to a stop?
(Unit = m)

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Answer

4. Solve for the distance up the hill (d): \( d = \frac{h}{\sin(10.0^{\circ})} \)

Steps

Step 1 ：1. Calculate the initial gravitational potential energy (PE): \( PE_{i} = mgh \)

Step 2 ：2. Calculate the final gravitational potential energy (PE): \( PE_{f} = PE_{i} - \frac{1}{2}mv_{i}^2 \)

Step 3 ：3. Solve for the height (h): \( h = \frac{PE_{f}}{mg} \)

Step 4 ：4. Solve for the distance up the hill (d): \( d = \frac{h}{\sin(10.0^{\circ})} \)